Use the Question 12 Writing Booklet (a) The vector $\mathbf{a}$ is $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and the vector $\mathbf{b}$ is $\begin{pmatrix} 2 \\ 0 \\ -4 \end{pmatrix}$ - HSC - SSCE Mathematics Extension 2 - Question 12 - 2024 - Paper 1

We need to show that the resulting vector is orthogonal to $\mathbf{b}$. We start from:

$\mathbf{c} = \mathbf{a} - \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}.$

Now to confirm that $\mathbf{c}$ is perpendicular to $\mathbf{b}$, we compute:

$\mathbf{c} \cdot \mathbf{b} = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 0 \\ -4 \end{pmatrix} = 2 \cdot 2 + 2 \cdot 0 + 1 \cdot (-4) = 4 + 0 - 4 = 0.$

Since the dot product is zero, this proves that $\mathbf{c}$ is perpendicular to $\mathbf{b}$.

To evaluate the integral, we first express the integrand using partial fractions:

$\frac{3x^2 + 2x + 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}.$

Multiplying both sides by $(x - 1)(x + 1)$ gives:

$3x^2 + 2x + 1 = A(x + 1) + B(x - 1).$

Expanding the right side:

$= Ax + A + Bx - B = (A + B)x + (A - B).$

Setting coefficients equal:

1. $A + B = 3$
2. $A - B = 1$

Solving this system:

• Adding the two equations gives $2A = 4$
• Therefore $A = 2$.
• Substituting $A$ into the first equation gives $2 + B = 3$, hence $B = 1$.

Now substituting back:

$\int \left( \frac{2}{x - 1} + \frac{1}{x + 1} \right) dx = 2 \ln |x - 1| + \ln |x + 1| + C.$

Given the equation $|z| = |z + 8 + 12i|$, we set $z = a + bi$. Thus, we have: $|a + bi| = |(a + 8) + (b + 12)i|.$

The magnitudes give: $\sqrt{a^2 + b^2} = \sqrt{(a + 8)^2 + (b + 12)^2}.$

Squaring both sides: $a^2 + b^2 = (a + 8)^2 + (b + 12)^2.$

Expanding the right side leads to: $a^2 + b^2 = a^2 + 16a + 64 + b^2 + 24b + 144.$

The terms $a^2$ and $b^2$ cancel: $0 = 16a + 24b + 208.$

Rearranging gives: $24b = -16a - 208 \Rightarrow b = -\frac{2}{3}a - \frac{208}{24} = -\frac{2}{3}a - \frac{52}{6}.$

Since $b$ must be real, plug in suitable values. Assuming $a = 0$, we find $b = -12$.

With $b = -12$, we can substitute back into $z$. So: $$z = a - 12i.$

To find $|z|$, we consider: $|z|^2 = a^2 + (-12)^2 = a^2 + 144.$

We also have: $|z + 8 + 12i|^2 = |(a + 8) + 0i|^2 = (a + 8)^2.$

Equating the two modulus equations: $a^2 + 144 = (a + 8)^2.$

This expands into: $a^2 + 144 = a^2 + 16a + 64.$

Therefore: $144 = 16a + 64 \implies 80 = 16a \implies a = 5.$

Thus, we find: $$z = 5 - 12i.$

First, observe the left-hand side $(n + 1)^{14} - 79n^{40}.$

For large values of $n$, $n^{40}$ dominates the expression.

• If $n$ is odd, $(n + 1)$ is even, so $ext{even}^{14}$ is even.
• $79n^{40}$ is odd for odd $n$.

This implies the whole expression is odd, hence cannot be 2 (even).

Conversely, if $n$ is even, $(n + 1)$ is odd.

Then, $ext{odd}^{14}$ is odd, while $79n^{40}$ remains even, making the whole expression odd again.

In either case, the left side cannot be 2, proving no integers satisfy the equation.

To find a vector equation for the line passing through points $A(3, 5, -4)$ and $B(7, 0, 2)$, we first calculate the direction vector:

$\mathbf{d} = \begin{pmatrix} 7 - 3 \\ 0 - 5 \\ 2 - (-4) \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \\ 6 \end{pmatrix}.$

The line equation can be presented as: $\ell: \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -5 \\ 6 \end{pmatrix}, \; \lambda \in \mathbb{R}.$

To check if $(10, 5, -2)$ lies on the line $\ell$, substitute into the line equation: $\begin{pmatrix} 10 \\ 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 + 4\lambda \\ 5 - 5\lambda \\ -4 + 6\lambda \end{pmatrix}.$

From this, we obtain three equations:

1. $10 = 3 + 4\lambda \Rightarrow 4\lambda = 7 \Rightarrow \lambda = \frac{7}{4}.$
2. $5 = 5 - 5\lambda \Rightarrow -5\lambda = 0 \Rightarrow \lambda = 0.$
3. $-2 = -4 + 6\lambda \Rightarrow 6\lambda = 2 \Rightarrow \lambda = \frac{1}{3}.$

Since we have inconsistent values for $\lambda$, the point $(10, 5, -2)$ does not lie on the line $\ell$.